login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A161170
Least integer k such that the n-almost prime count is equal to the prime count.
0
10, 125, 1809, 37820, 2722768, 1037849736, 4204496515890
OFFSET
2,1
COMMENTS
Related to sequence A125149, but we compare the prime count to the semiprime count, then the product-of-three-primes count, and so on.
We start with a the number two, and a prime count of 1.
Then the prime count and semiprime count are first identical when k = 10, the prime count is 4 ({2, 3, 5, 7}) and the semiprime count is also 4 ({4, 6, 9, 10}).
Next, when k = 125 the prime count of 30 and product-of-three-primes count of 30 are first identical.
Based on the write up for A125149 and examination of the factor counts as k increases, we believe this sequence is infinite, but have not proved this.
EXAMPLE
a(2) = 10 since there are now 4 primes ({2, 3, 5, 7}) and 4 semiprimes ({4, 6, 9, 10}) <= 10.
a(3) = 125 with 30 primes and 30 products of 3 primes.
a(4) = 1809 with 279 primes and 279 products of 4 primes.
a(5) = 37820 with 4000 primes and 4000 products of 5 primes.
a(6) = 2722768 with 198183 primes and 198183 products of 6 primes.
a(7) = 1037849736 with 52672391 primes and 52672391 products of 7 primes.
a(8) = 4204496515890 with 150007470826 primes and 150007470826 products of 8 primes.
PROG
(Ruby) # A slow program to generate sequence
# Faster C code is available by request
# Tallies number of primes, semiprimes, trieneprimes ...
tally = Hash.new { |h, k| h[k] = 0}
# Returns number of factors of num
def factors(num)
(2..(Math.sqrt(num).to_i)).each{ |i|
return factors(num/i) + 1 if num % i == 0
}
1
end
# Testing number of primes against composites with num_factors
num_factors = 2
2.upto( 1.0/0.0 ) { |i|
tally[factors(i)] +=1
if tally[1] == tally[num_factors]
puts "k: #{i} Primes: #{tally[1]} Composites with #{num_factors} factors: #{tally[num_factors]}"
num_factors += 1
end
}
(Perl) use ntheory ":all"; my($k, @S)=(0, map{0}1..20); forfactored { $S[@_]++; while ($S[1] == $S[$k]) { print "$k $_\n" if $k>1; $k++; } } 3e6; # Dana Jacobsen, Jan 18 2019
CROSSREFS
Cf. A125149.
Sequence in context: A123358 A230390 A089832 * A281595 A097816 A323877
KEYWORD
hard,more,nonn
AUTHOR
Andy Martin, Jun 04 2009
EXTENSIONS
Edited example and a(8) from Donovan Johnson, Mar 10 2010
STATUS
approved